Q11.After the collision, how do I know which velocity is greater? If v1 and v2 are the respective velocities:
1) If v1>v2, I get v1 = 30, v2 = 21
S = 21 * 5 = 105
105 = v1 * t1
t1 = 3.5 s
2) If v2 > v1, I get v1 = 15, v2 = 24
S = 24 * 5 = 120
120 = v1 * t1
t1 = 6 s
The book gives 3 s as the answer. And in general, if I didn't know that the first particle will meet the second one after the collision, shouldn't I also consider the case when the first particle changes direction? What am I missing?
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 30042016 12:57

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 30042016 14:51
(Original post by to4ka)
Q11.After the collision, how do I know which velocity is greater? If v1 and v2 are the respective velocities:
1) If v1>v2, I get v1 = 30, v2 = 21
S = 21 * 5 = 105
105 = v1 * t1
t1 = 3.5 s
2) If v2 > v1, I get v1 = 15, v2 = 24
S = 24 * 5 = 120
120 = v1 * t1
t1 = 6 s
The book gives 3 s as the answer. And in general, if I didn't know that the first particle will meet the second one after the collision, shouldn't I also consider the case when the first particle changes direction? What am I missing?
Does that cover your misunderstanding?
Edit: There is only one possible set of values for v1,v2. How come you have two? EndEdit.
You also made a siip at the end "120 = v1 * t1" implies v1 = ?? Not 6.
And note that they want the additional time, not just the time since the impact.Last edited by ghostwalker; 30042016 at 14:56. 
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 30042016 14:51
I just realized v1 > v2 makes no sense. And if the first particle changes direction, energy is created... but shouldn't the two cases be discussed when solving such problems?
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 30042016 15:00
(Original post by ghostwalker)
The text of the question says 2kg and 10 kg are travelling in the same direction initially. The 2kg mass catches up with the 10 kg mass, and they collide.
Does that cover your misunderstanding?
You also made a siip at the end "120 = v1 * t1" implies v1 = ?? Not 6.
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 30042016 15:02
(Original post by to4ka)
Does "collide" imply they don't change directions?Last edited by ghostwalker; 30042016 at 15:04. 
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 30042016 15:08
And what about the case when A changes direction? I know it's impossible in this question as you're told they collide again, but if the question was shorter and asked for the final velocities only, shouldn't we consider the case?
Because, if A changes direction:
e = (v1 + v2) / 15
and since by cons. of momentum
270 = v1 * 2 + 10 * v2
I get
v2 = 31,5
v1 =  22.5
which is impossible as spd is positive. But why is this case ignored? Shouldn't I prove that the particles don't change direction of motion? 
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 30042016 15:10
(Original post by to4ka)
I just realized v1 > v2 makes no sense. And if the first particle changes direction, energy is created... but shouldn't the two cases be discussed when solving such problems?
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You can tell from the wording of the question that v1,v2 are both going to be in the same direction as previous.
If you inadvertently defined v1 in the opposite direction, it's not a problem as the value you'd get would be negative showing that the direction is actually opposite to the one you assumed. 
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 30042016 15:12
Your equations do not tell you particles don't change direction  you assume it by writing
e = (v2  v1) / 15
because if they change direction, this becomes
e = (v2 + v1) / 15
EDIT: I define v1 and v2 as spd rather than velocity. If i define them as velocities, I can't know if the relative speed after the collision is v2  v1 or v1 + v2.
Posted from TSR MobileLast edited by to4ka; 30042016 at 15:14. 
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 30042016 15:22
(Original post by to4ka)
Your equations do not tell you particles don't change direction  you assume it by writing
e = (v2  v1) / 15
because if they change direction, this becomes
e = (v2 + v1) / 15
EDIT: I define v1 and v2 as spd rather than velocity. If i define them as velocities, I can't know if the relative speed after the collision is v2  v1 or v1 + v2.
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 30042016 15:40
(Original post by to4ka)
Your equations do not tell you particles don't change direction  you assume it by writing
e = (v2  v1) / 15
because if they change direction, this becomes
e = (v2 + v1) / 15
EDIT: I define v1 and v2 as spd rather than velocity. If i define them as velocities, I can't know if the relative speed after the collision is v2  v1 or v1 + v2.
Posted from TSR Mobile
If you write v2  v1, then your answer will be (let's say) v_1 = 20 and v_2 = 30.
had you written v_2 + v_1, your answer would be v_1 = 20 and v_2 = 30.
In both cases, you'd have gotten 30  20 = 10.
i.e: just use one set and the algebra will sort itself out and give you a positive/negative value depending on whether your assumed direction is the same as the actual direction. If your answer is positive, the assumed direction is the same as the actual direction. If your answer is negative, the assumed direction is the opposite of the actual direction. 
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 30042016 16:19
(Original post by Zacken)
You can't "define v1 and v2 as speed"  you are using them in a momentum equation which is a vector equation (momentum is a vector quantity).
If you write v2  v1, then your answer will be (let's say) v_1 = 20 and v_2 = 30.
had you written v_2 + v_1, your answer would be v_1 = 20 and v_2 = 30.
In both cases, you'd have gotten 30  20 = 10.
i.e: just use one set and the algebra will sort itself out and give you a positive/negative value depending on whether your assumed direction is the same as the actual direction. If your answer is positive, the assumed direction is the same as the actual direction. If your answer is negative, the assumed direction is the opposite of the actual direction.
Let (>) be positive, define v1 and v2 as the final velocities of A and B.
By conservation of momentum, we have:
35 * 2 + 20 * 10 = v1 * 2 + v2 * 10
so
v1 = 135  5v2
We also have that
e = (relative speed of separation) / (relative speed of approach) = ??? / 15
What is ???  surely it can't be (v2  v1) as this is relative velocity; I need the spd. It must be v2v1 = 6v2  135
which gives
6v2135 = 9
so v2 is either 21 or 24. In the first case I get v1 = 30 > v1 which is impossible and the second case is all good. Is this right? Or is there a simpler way?
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 30042016 16:22
(Original post by to4ka)
Right, this is all good, but I get stuck in the middle:
Let (> be positive, define v1 and v2 as the final velocities of A and B.
By conservation of momentum, we have:
35 * 2 + 20 * 10 = v1 * 2 + v2 * 10
so
v1 = 135  5v2
We also have that
e = (relative speed of separation) / (relative speed of approach) = ??? / 15
What is ???  surely it can't be (v2  v1) as this is relative velocity; I need the spd. It must be v2v1 = 6v2  135
which gives
6v2135 = 9
so v2 is either 21 or 24. In the first case I get v1 = 30 > v1 which is impossible and the second case is all good. Is this right? Or is there a simpler way?
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So, indeed, it is (v2  v1)/15. 
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 30042016 16:35
Uhm, not? Relative velocity is a vector and you can't divide vectors (not at this point, at least) by vectors.
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 30042016 16:38
(Original post by to4ka)
Uhm, not? Relative velocity is a vector and you can't divide vectors (not at this point, at least) by vectors.
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 30042016 16:45
Techincally,
time = displacement/velocity
as in, if velocity is 3i+4j and displacement is 15j
time isn't (15j)/(3i+4j), it's 15 / sqrt(3*3+4*4), which is 3.
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 30042016 16:50

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 30042016 16:53
(Original post by to4ka)
Techincally,
time = displacement/velocity
as in, if velocity is 3i+4j and displacement is 15j
time isn't (15j)/(3i+4j), it's 15 / sqrt(3*3+4*4), which is 3.
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It's just that relative speed = difference in velocities. 
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 30042016 17:00
Are you sure?
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 01052016 13:33
(Original post by to4ka)
...
Two cases:
1. We take v1 and v2 going in the same direction as the initial velocities.
By momentum we have: and so:
By restitution:
"speed separation" = e( "speed of approach" )
Hence,
(1)+(2) gives
And thus , and subbing into (2) we get
2. We take v1 going back in the opposite diretion and v2 going in the same direction as the intial velocities.
By momentum we have: and so:
By restitution:
"speed separation" = e( "speed of approach" )
Hence,
(3)+(4) gives
And thus , and subbing into (4) we get
Since we assumed v1 was going back in the opposite direction, and we get a minus value, then v1 is in fact going in the original direction, same as all the other velocities.Last edited by ghostwalker; 01052016 at 13:45.Post rating:1 
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 01052016 14:38
Thanks, I understood quite a lot from you both!
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Updated: May 1, 2016
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